Projectile Motion, speed of ball

  • Thread starter Lara
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  • #1
I cant seem to get the right answer to this question so if someone could guide me through, please?

A ball is shot at a goal from a horizontal distance of 5.3m. The ball is released at an angle of 48 degrees to the horizontal and 1.2m below the height of the goal. What is the ballls speed?

thankyou
 

Answers and Replies

  • #2
Integral
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Did you give us all the information you have? I see no way, with the given information, to specify a single speed. Do you want the speed required to hit the goal?
 
  • #3
Integral
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Assuming that you only need the projectile to be at the goal (5.3,1.2) you can get the speed with the following equations.

1. The components of the speed are Vx=VCosΘ and
Vy=VSinΘ

V is the speed, Θ=48deg

2. x(t) = Vxt with the condition that x(T)=5.3

3. y(t)= -gt2/2 + Vyt + Y with conditions y(0)=0 and Y(T)=1.2

This should be a nice start for you.
 

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